Question

I'm beginning to learn cohomology for cyclic groups in preparation for use in the proofs of global class field theory (using ideals). I've seen the proof of the long exact sequence and of basic properties of the Herbrand quotient, and I've started to look through how it's used in the proofs of class field theory. So far, all I can tell is that the cohomology groups are given by some random modding out process, and then we derive some random properties (like the long exact sequence), and then we compute things like $H^0(I_{L},\mathrm{Gal}(L/K))$, where $I_L$ denotes the group of fractional ideals of a number field L, and it happens to be something interesting for the study of class field theory $(I_K/N(I_L))$, where $L/K$ is cyclic and $N$ denotes the ideal norm). We then find that the cohomology groups are useful for streamlining the computations with various orders of indexes of groups. What I don't get is what the intuition is behind the definitions of these cohomology groups. I do know what cohomology is in a geometric setting (so I know examples where taking the kernel mod the image is interesting), but that doesn't help here. What is the intuition for why they are defined the way they are? Why should we expect that these cohomology groups so-defined have nice properties and help us with algebraic number theory? Right now, I just see theorem after theorem, I see the algebraic manipulation and diagram chasing that proves it, but I don't see a bigger picture.

For context, if A is a G-module where G is cyclic and $\sigma$ is the generator of G, then we define the endomorphisms $D=1+\sigma+\sigma^2+...$ and $N=1-\sigma of A$, and then $H^0=kerN/imD$ and $H^1=kerD/imN$.

Edit: The thread http://mathoverflow.net/questions/8599/tips-on-cohomology-for-number-theory posted be Leonid Postiselski is very pertinent and useful!

Answer 1

Here is a completely elementary example which shows that group cohomology is not empty verbiage, but can solve a problem ("parametrization of rational circle") whose statement has nothing to do with cohomology.

Suppose you somehow know that for a finite Galois extension $k\subset K$ with group $G$ the first cohomology group $H^1(G,K^*)$ is zero : this is the homological version of Hilbert's Theorem 90 ( you can look it up in Weibel's book on homological algebra, pages 175-176).

If moreover $G $ is cyclic with generator $s$, this implies that an element of $K$ has norm one if and only if it can be written $\frac{a}{s(a)}$ for some $a\in K$.

Consider now the quadratic extension $k=\mathbb Q \subset K=\mathbb Q(i)$ with generator $s$ of $Gal(\mathbb Q (i)/\mathbb Q)$ the complex conjugation.The statement above says that $x+iy\in \mathbb Q(i)$ satisfies $x^2+y^2=1$ iff $x+iy=\frac{u+iv}{s(u+iv)}=\frac{u+iv}{u-iv}=\frac{u^2-v^2}{u^2+v^2}+i\frac{2uv}{u^2+v^2}$ for some $u+iv\in \mathbb Q (i)$ .

So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle $x^2+y^2=1$ $$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$.

Answer 2

I'm not sure if this is what you're looking for, but I always think of group (co)homology in terms of the homology of the classifying space for your group. Assuming $G$ is discrete, then there is a topological space $BG$ with the property that $\pi_1 BG=G$ and the higher homotopy groups vanish. By construction, $BG$ has a contractible cover $EG$ so that $EG/G=BG$.

$H_n(BG)$ is the same as the algebraically defined $H_n(G)$ since, if we take the cellular chain complex of $EG$ we end up with a resolution of the integers by $G$-modules because of the action of $G$ on $EG$ passes to the chain groups. Then tensoring by the integral group ring of $G$ just divides out the $G$ action and we get the cellular chain complex of $BG$.

Answer 3

I feel like my answer to every question of this kind is "the treatment in Silverman's book on elliptic curves is really nice," but the treatment in Silverman's book on elliptic curves is really nice!

In particular, for H^1 at least I always find it quite nice to think in terms of twists. Quite generally: if X is a variety over a field K, and L/K is a Galois extension, we say that X'/K is a L-twist of X if there exists an isomorphism between X and X' over L. (If we just mean X' is an L-twist for some L, we just call it a twist of X.)

Anyway, it is a nice exercise to check that L-twists of X yield classes in H^1(Gal(L/K),Aut(X/L)). (And in favorable circumstances, the L-twists are actually in bijection with the Galois cohomology set.) This is the basis of the whole story of principal homogeneous spaces, or torsors, which makes up much of the last chapter of Silverman.

Answer 4

The online notes of J.S. Milne on Class Field Theory contain a chapter on group cohomology including Tate cohomology that is easily accessible (and exactly what you need to understand class field theory).

For algebraic intuition the keyword is derived functors. Group cohomology (and homology) are examples of derived functors, which can be considered as a reason why the definitions as are they are, and why they are interesting/useful.

Answer 5

First, let me say beyond knowing that group cohomology comes from derived functors and all the properties that come with it, I don't have much intuition for general group cohomology. However, in number theory there are a number of places in which you can realize the cohomology groups as parametrizing some other objects you are interested in. To understand these other objects (which you could feasibly have very real intuition about) it then suffices to use the machinery of group cohomology. Below are a few examples. I forget to put continuous subscripts on everything, so beware. Mistakes are mine.

First, in the realm of (edit:local) class field theory we have the Brauer group $Br(K)$ which is the abelian group (under the tensor operation) of central simple algebras over $K$ up to the equivalence $A \sim M_n(A)$. It turns out that there is a natural isomorphism between $Br(K)$ and $H^2(G_K, \overline{K}^{\times})$. This isomorphism provides two worlds in which to make important calculations. For instance, the statement that every central simple algebra over $K$ splits over an unramified extension of $K$ may be proved explicitly using the Brauer group or it may be proven by checking that $H^2(G(K^{nr}/K),\overline{K}^\times) = H^2(G_K,\overline{K}^\times)$. From either proof you are able to obtain a proof of the other. Perhaps, also in the realm of class field theory, the local Artin map arises as a map on Tate groups given by a certain cup product but that would take a bit longer to explain. You should look at Milne's notes on CFT for all of this (Ch III and Ch IV for what I have said).

Here is another. Suppose that $B$ is a topological ring and $G$ is a topological group and $G$ acts continuously on $B$. Then, we consider finite free $B$-modules $X$ equipped with a semi-linear action, i.e. $g(bx) = g(b)g(x)$ for all $b \in B$ and $x \in X$. It turns out that all such objects are parametrized by $H^1(G,GL_d(B))$. (Warning: this is non-abelian cohomology, so this is only a pointed set. The point corresponds to the trivial semi-linear representation $B^d$ with the diagonal action.) This comes up very early in the part of $p$-adic Galois representations where one studies the period rings $B_{dR}, B_{HT}$, etc.

Finally, consider the situation where one has a representation $\overline{\rho}:G_{\mathbb Q,S} \rightarrow GL_n(\mathbb F_p)$ and one wants to know whether it has litings $\rho: G_{\mathbb Q,S} \rightarrow GL_n(R)$ where $R$ is some complete DVR with residue field $\mathbb F_p$. If we already have a lifting to $GL_n(R')$ and $R$ and $R'$ are nice enough (there is a surjection $R \rightarrow R'$ whose kernel $I$ is killed by the maximal ideal in $R$) then the obstruction to lifting further lies in the cohomology group $H^2(G_{\mathbb Q,S},I\otimes Ad(\overline{\rho}))$ where $Ad(\overline{\rho})$ is the vectorspace $M_n(\mathbb F_p)$ together with conjugate action by $\overline{\rho}$. This (I've specialized a few things) is written down in Mazur's paper "Deforming Galois Representations".

To finish, I will just say what my adviser told me when I asked him a similar question as you are asking: "Just wait and you will see how much clearer your thought can become with group cohomology."

Answer 6

In my opinion, the best you can do is to see group cohomology in the context of homological algebra. For example, the book by Hilton and Stammbach on homological algebra should be a good introduction.